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Unformatted text preview: Math 2374 Spring 2006 Midterm 2 Solution  Page 1 of 5 March 29, 2006 1. (25 points) Let the curve C in the ( x, y )plane be the boundary of the unit square: C consists of four line segments, from (0 , 0) to (1 , 0); from (1 , 0) to (1 , 1); from (1 , 1) to (0 , 1); and from (0 , 1) to (0 , 0). Evaluate the line integral C xy ( 1 + x 2 + 9) dx + 1 3 ( x 2 + 9) 3 / 2 dy by using Green’s Theorem. Solution. Need to observe first that C is a closed curve which is the boundary of the unit square D : x 1, 0 y 1. Since it is closed we may use Green’s theorem: C P dx + Q dy = D ∂Q ∂x ∂P ∂y dxdy In our case P = xy ( x 2 + 9 1), ∂P ∂y = x ( x 2 + 9 1) = x x 2 + 9 x Q = 1 3 ( x 2 + 9) 3 / 2 , ∂Q ∂x = 1 3 3 2 2 x ( x 2 + 9) 1 / 2 = x x 2 + 9 Thus ∂Q ∂x ∂P ∂y = x x 2 + 9 ( x x 2 + 9 x ) = x By Green’s theorem C P dx + Q dy = D x dxdy = 1 1 x dy dx = 1 x dx = 1 / 2 . Deductions. Mistake in the computation of the last integral 2 pts off. Wrong computation of ∂Q ∂x ∂P ∂y leads to 57 pts off provided that integration after still done up to a ”right” number. If not, other deductions might take place. Math 2374 Spring 2006 Midterm 2 Solution  Page 2 of 5 March 29, 2006 2. (25 points) You build a fence so that the base of the fence is the circle C given by x 2 + y 2 = 4....
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This note was uploaded on 04/07/2008 for the course MATH 2374 taught by Professor Mosher during the Spring '07 term at Minnesota.
 Spring '07
 Mosher
 Math

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